Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates

Question

How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$ when $E$ is bounded by $x^2+y^2+z^2-x=0$?

I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$ and this function in spherical coordinates is \begin{align*} &\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi \end{align*} but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.

Answer 1

The set $E$ is a ball of radius ${1\over2}$, centered at $\left({1\over2},0,0\right)$. The integrand $$f(x,y,z):={yz\over x^2+y^2+z^2}$$ satisfies $f(x,-y,z)\equiv-f(x,y,z)$. This means that $f$ is odd with respect to the symmetry plane $y=0$ of $E$. It is then obvious that the requested integral has value $0$.

Answer 2

Using the spherical coordinates (with the convention you used) we have, $$yz= r^2\sin\theta\cos\theta\sin\varphi,$$ $$x^2+y^2+z^2 = r^2$$ $$\,dx\,dy\,dz = r^2\sin\theta\,dr\,d\varphi\,d\theta.$$ Also, the region $E$ is bounded by $x^2+y^2+z^2-x = 0$, which is a sphere centered at $(\tfrac{1}{2},0,0)$ with radius $\tfrac{1}{2}$.

The inside of this sphere is given by $x^2+y^2+z^2-x \le 0$, i.e. $r^2-r\sin\theta\cos\varphi \le 0$. Since $r \ge 0$, this simplifies to $0 \le r \le \sin\theta \cos\varphi$ or $r = 0$ (which is just one point).

In order for $0 \le r \le \sin\theta\cos\varphi$ to be a nontrivial range, we need $\sin\theta\cos\varphi \ge 0$. Since $\sin\theta \ge 0$ for all $\theta \in [0,\pi]$, we only need to restrict the bounds of $\varphi$ to be over the range where $\cos\varphi \ge 0$. We can do this either by $\varphi \in [0,\tfrac{\pi}{2}] \cup [\tfrac{3\pi}{2},2\pi]$ or $\varphi \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. The second is simpler, so we will use that instead.

So the integral becomes $$\iiint\limits_{E}\dfrac{yz \,dx\,dy\,dz}{x^2+y^2+z^2} = \int_{-\pi/2}^{\pi/2}\int_{0}^{\pi}\int_{0}^{\sin\theta\cos\varphi}r^2\sin^2\theta\cos\theta\sin\varphi\,dr\,d\theta\,d\varphi.$$

Answer 3

$x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$

So it is a sphere with radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.

In spherical coordinates, $x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$ where $\theta$ is the azimuthal angle and $\phi$ is the polar angle (just opposite of your convention).

$x^2 + y^2 + z^2 = \rho^2$ so substituting in the equation of our sphere,

$\rho^2 = \rho \cos \theta \sin \phi \implies \rho = \cos \theta \sin \phi$.

Now the part that you have to be careful about is the bounds of azimuthal angle. Please consider a circle in XY plane with center at $(\frac{1}{2},0)$ and radius of $\frac{1}{2}$ and equation of $r = \cos \theta$. $\theta$ varies betweeen -$\frac{\pi}{2}$ and $\frac{\pi}{2}$ (range of $\pi$ instead of $2 \pi$ for a sphere centered at the origin). It is same for this sphere.

So the integral becomes

$I = \displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta$

And based on the given function, its integral above $XY$ plane and below will cancel each other.

If I was integrating over the part of the sphere only in the first octant, the integral will be

$I = \displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta = \frac{1}{72}$